Asked by Thalia
if f(x)=ln(x^3 +1) find f'(3), correct to 2 decimal places
Answers
Answered by
drwls
Use the chain rule, with u = x^3 + 1
f(x) = ln u(x)
df/dx =df/du*du/dx
= 1/u * 3x^2
= 3x^2/[x^3 +1]
Now plug in x = 3 and turn the crank.
f(x) = ln u(x)
df/dx =df/du*du/dx
= 1/u * 3x^2
= 3x^2/[x^3 +1]
Now plug in x = 3 and turn the crank.
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